String

14. Word Ladder

1. The problem is called "Word Ladder" and the problem link on LeetCode is: https://leetcode.com/problems/word-ladder/

2. Here is the optimized Java code for the "Word Ladder" problem with detailed comments:

import java.util.*;

public class WordLadder {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        // Create a set to store all the words in the word list
        Set<String> wordSet = new HashSet<>(wordList);

        // If the end word is not in the word list, there is no path
        if (!wordSet.contains(endWord)) {
            return 0;
        }

        // Create a queue for BFS and add the begin word to it
        Queue<String> queue = new LinkedList<>();
        queue.offer(beginWord);

        // Create a set to track the visited words
        Set<String> visited = new HashSet<>();
        visited.add(beginWord);

        // Start the BFS with level 1
        int level = 1;

        while (!queue.isEmpty()) {
            int size = queue.size();

            // Process all the words at the current level
            for (int i = 0; i < size; i++) {
                String currentWord = queue.poll();

                // Check all possible transformations of the current word
                for (int j = 0; j < currentWord.length(); j++) {
                    char[] wordArray = currentWord.toCharArray();

                    // Try replacing each character in the word with a-z
                    for (char c = 'a'; c <= 'z'; c++) {
                        wordArray[j] = c;
                        String transformedWord = new String(wordArray);

                        // If the transformed word is the end word, we found the shortest path
                        if (transformedWord.equals(endWord)) {
                            return level + 1;
                        }

                        // If the transformed word is in the word set and not visited, add it to the queue and set
                        if (wordSet.contains(transformedWord) && !visited.contains(transformedWord)) {
                            queue.offer(transformedWord);
                            visited.add(transformedWord);
                        }
                    }
                }
            }

            // Increase the level after processing all words at the current level
            level++;
        }

        // If we reach here, there is no path from begin word to end word
        return 0;
    }
}

1. Algorithm and Interview Approach:

• The "Word Ladder" problem can be solved using a breadth-first search (BFS) approach.

• The basic idea is to start from the given begin word and explore all possible transformations by replacing one character at a time with a-z.

• We use a queue to perform the BFS. Each level of the BFS represents a transformation of the current word.

• We also use a visited set to keep track of the visited words to avoid cycles in the graph.

• At each level, we check all possible transformations of the current word and add them to the queue if they are in the word list and not yet visited.

• We continue the BFS until we find the end word or there are no more words to explore.

• If we find the end word, we return the level of the BFS as the shortest transformation path.

• If we reach the end of the BFS without finding the end word, there is no path from the begin word to the end word.

• During an interview, it's important to first understand the problem statement and clarify any ambiguities.

• Discuss the possible approaches and their time complexities.

• If a BFS solution is chosen, explain the algorithm and its steps in detail, including the data structures used.

• Walk through an example to make sure the logic is clear.

• Implement the code with proper variable names and comments to demonstrate your coding skills and understanding of the problem.

• Test the code with different test cases to validate its correctness and efficiency.

15. Word Ladder II

1. Problem Description:

The problem "Word Ladder II" on LeetCode is a popular problem that tests your understanding of graph traversal and path finding algorithms. Given two words, beginWord and endWord, and a word list, find all shortest transformation sequences from beginWord to endWord, such that:

1. Each transformation sequence must begin with beginWord and end with endWord.

2. Each transformation must differ by only one letter.

3. Each transformation must use only words from the word list.

You can assume that:

1. The word list contains only lowercase alphabets.

2. The length of beginWord and endWord is the same.

3. beginWord and endWord are not the same.

Here is the link to the problem on LeetCode: https://leetcode.com/problems/word-ladder-ii/

1. Optimized Java Code with Detailed Comments:

import java.util.*;

public class Solution {
    public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
        List<List<String>> result = new ArrayList<>();
        
        // Convert wordList to Set for faster lookup
        Set<String> wordSet = new HashSet<>(wordList);
        
        // If endWord is not in wordList, it is impossible to reach
        if (!wordSet.contains(endWord)) {
            return result;
        }
        
        // Using a map to store the parent-child relationship between words
        Map<String, List<String>> graph = new HashMap<>();
        Set<String> visited = new HashSet<>();
        Set<String> currentLevel = new HashSet<>();
        Set<String> nextLevel;
        
        currentLevel.add(beginWord);
        boolean found = false;
        
        while (!currentLevel.isEmpty() && !found) {
            // Mark words in currentLevel as visited
            visited.addAll(currentLevel);
            
            // Stores the words for the next level
            nextLevel = new HashSet<>();
            
            for (String word : currentLevel) {
                // Generate all possible next words by changing one letter
                char[] chars = word.toCharArray();
                for (int i = 0; i < chars.length; i++) {
                    char originalChar = chars[i];
                    for (char c = 'a'; c <= 'z'; c++) {
                        chars[i] = c;
                        String nextWord = new String(chars);
                        
                        if (wordSet.contains(nextWord)) {
                            // Create a parent-child relationship between words
                            if (!graph.containsKey(nextWord)) {
                                graph.put(nextWord, new ArrayList<>());
                            }
                            graph.get(nextWord).add(word);
                            
                            if (nextWord.equals(endWord)) {
                                // Reach the end word, stop exploring
                                found = true;
                            } else if (!visited.contains(nextWord)) {
                                // Add valid nextWord to the next level
                                nextLevel.add(nextWord);
                            }
                        }
                    }
                    chars[i] = originalChar;
                }
            }
            
            // Move to the next level
            currentLevel = nextLevel;
        }
        
        if (found) {
            // Backtracking to construct all possible transformation sequences
            List<String> path = new ArrayList<>();
            path.add(endWord);
            buildPath(result, graph, path, endWord, beginWord);
        }
        
        return result;
    }
    
    private void buildPath(List<List<String>> result, Map<String, List<String>> graph, List<String> path, String word, String beginWord) {
        if (word.equals(beginWord)) {
            // Reverse the path since it was constructed from endWord to beginWord
            Collections.reverse(path);
            result.add(new ArrayList<>(path));
            Collections.reverse(path);
            return;
        }
        
        if (graph.containsKey(word)) {
            for (String parent : graph.get(word)) {
                path.add(parent);
                buildPath(result, graph, path, parent, beginWord);
                // Backtracking
                path.remove(path.size() - 1);
            }
        }
    }
}

1. Algorithm and Approach:

To solve this problem efficiently, we can use a breadth-first search (BFS) approach. The idea is to treat the word ladder as a graph, where each word is a node and there is an edge between two words if they differ by only one letter. The goal is to find the shortest path from the beginWord to the endWord in this graph.

Here are the steps to follow during an interview process:

1. Convert the wordList into a Set data structure for fast lookup. This will help in determining if a word is present in the wordList in constant time.

2. Create a graph using HashMap, where the key is the word and the value is a list of its parent words (words that can be transformed into the current word).

3. Initialize a visited set, a currentLevel set (containing the words to be explored in the current level), and a found flag (to track if we have reached the endWord).

4. Start with the beginWord in the currentLevel set.

5. While the currentLevel set is not empty and the endWord is not found:

   • Mark the words in currentLevel as visited.

   • Create the nextLevel set to store the words for the next level of exploration.

   • For each word in the currentLevel set, generate all possible next words by changing one letter at a time.

   • If the nextWord is present in the wordSet and has not been visited before, create a parent-child relationship between the current word and the nextWord in the graph.

   • If the nextWord is the endWord, set the found flag to true.

   • Add every valid nextWord to the nextLevel set.

   • Move to the next level by assigning nextLevel to currentLevel.

6. If the endWord is found, perform backtracking to construct all possible transformation sequences from the endWord to the beginWord using the parent-child relationship stored in the graph.

7. Return the final result, which is a list of all possible transformation sequences.

Understanding and implementing this algorithm is crucial in solving similar graph traversal and path finding problems during an interview. It demonstrates your ability to think critically and efficiently solve complex problems.

16. Number of Matching Subsequences

1. Problem Description: The problem "Number of Matching Subsequences" on LeetCode asks us to find the number of words in a given string array that are subsequences of another string. Here is the link to the problem: https://leetcode.com/problems/number-of-matching-subsequences/

2. Optimized Java Code with detailed comments:

public class NumberofMatchingSubsequences {
    public int numMatchingSubseq(String s, String[] words) {
        int count = 0;
        
        // Create a list for each character in the alphabet
        ArrayList<Integer>[] list = new ArrayList[26];
        for (int i = 0; i < 26; i++) {
            list[i] = new ArrayList<>();
        }
        
        // Iterate through the main string and add the indices of each character to the corresponding list
        for (int i = 0; i < s.length(); i++) {
            list[s.charAt(i) - 'a'].add(i);
        }
        
        // Iterate through the word array
        for (String word : words) {
            int currIndex = -1;
            boolean isSubsequence = true;
            
            // Iterate through each character of the word
            for (char ch : word.toCharArray()) {
                ArrayList<Integer> tempList = list[ch - 'a'];
                int lowerBound = binarySearch(tempList, currIndex);
                
                // If the character is not found or is found at an earlier index, word is not a subsequence
                if (lowerBound == tempList.size() || tempList.get(lowerBound) <= currIndex) {
                    isSubsequence = false;
                    break;
                }
                
                currIndex = tempList.get(lowerBound);
            }
            
            // If word is a subsequence, increment count
            if (isSubsequence) {
                count++;
            }
        }
        
        return count;
    }
    
    // Binary search to find the index of the smallest element greater than the current index
    private int binarySearch(ArrayList<Integer> list, int target) {
        int left = 0;
        int right = list.size() - 1;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            
            if (list.get(mid) <= target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        return left;
    }
}

1. Algorithm and Interview Process:

The key to solving this problem efficiently is to pre-process the main string and store the indices of each character separately to avoid searching for each character multiple times. This can be accomplished by creating an array of lists, where each list represents the indices of a specific character.

The algorithm then iterates through each word in the array and checks if each character in the word is found in the corresponding list. To optimize the process of finding the next index of a character, we use binary search on the list to find the index of the smallest element greater than the current index. If we find the character at an earlier index or cannot find it at all, we know that the word is not a subsequence.

During an interview process, you can follow these steps to tackle this problem efficiently:

1. Understand the problem statement and requirements.

2. Recognize the need to optimize the searching process by pre-processing the main string.

3. Think about data structures that can be employed to improve efficiency.

4. Consider using an array of lists to store the indices of each character.

5. Implement the code step by step, explaining your thought process and adding comments to clarify each step.

6. Optimize the searching process by using binary search to find the next index of a character.

7. Test the code with different test cases and discuss its time and space complexity.

8. Discuss possible improvements or alternative approaches if time permits.

By thoroughly understanding the problem, explaining the algorithm clearly, and optimizing the code, you can demonstrate your problem-solving skills and interview readiness.

17. As Far from Land as Possible

1. Problem Description: The problem "As Far from Land as Possible" on LeetCode is described as follows:

Given an n x n grid containing only values 0 and 1, where 0 represents water and 1 represents land, find the maximum distance from water to land. The distance is calculated using Manhattan distance, where distance(a, b) = |a.x - b.x| + |a.y - b.y|. If there is no land or water in the grid, return -1.

LeetCode link: https://leetcode.com/problems/as-far-from-land-as-possible/

1. Optimized Java Code with Detailed Comments:

class Solution {
    public int maxDistance(int[][] grid) {
        // Initialize a queue to hold the coordinates of land cells
        Queue<int[]> queue = new LinkedList<>();
        int n = grid.length;
        
        // Add all land cells to the queue
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    queue.offer(new int[]{i, j});
                }
            }
        }
        
        // Check if there are no water or land cells
        if (queue.isEmpty() || queue.size() == n * n) {
            return -1;
        }
        
        int[][] directions = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        int distance = -1;
        
        // Process all land cells to find the maximum distance from water
        while (!queue.isEmpty()) {
            distance++;
            int size = queue.size();
            
            for (int k = 0; k < size; k++) {
                int[] cell = queue.poll();
                int x = cell[0];
                int y = cell[1];
                
                for (int[] dir : directions) {
                    int newX = x + dir[0];
                    int newY = y + dir[1];
                    
                    // Check if the adjacent cell is within the grid boundaries and is water
                    if (newX >= 0 && newX < n && newY >= 0 && newY < n && grid[newX][newY] == 0) {
                        grid[newX][newY] = 1; // Mark the adjacent cell as land
                        queue.offer(new int[]{newX, newY}); // Add the adjacent cell to the queue for further processing
                    }
                }
            }
        }
        
        return distance;
    }
}

1. Detailed Description of the Algorithm and Interview Approach:

The problem can be solved using a Breadth-First Search (BFS) approach. Here is the algorithm:

1. Create a queue to hold the coordinates of the land cells.

2. Traverse the entire grid and add the coordinates of each land cell to the queue.

3. Check if there are no land or water cells in the grid. If so, return -1.

4. Create a 2D array "directions" to store the possible directions to move - up, down, left, and right.

5. Initialize a variable "distance" to -1 to keep track of the maximum distance from water.

6. Process each land cell in the queue:

   • Initialize a variable "size" to keep track of the number of land cells in the current level of BFS.

   • Iterate over each land cell in the current level:

     • Get the coordinates (x, y) of the land cell.

     • Iterate over the possible directions and calculate the coordinates of the adjacent cells.

     • Check if the adjacent cell is within the grid boundaries and is water (grid[newX][newY] == 0).

       • If so, mark that cell as land (grid[newX][newY] = 1) and add its coordinates to the queue for further processing.

   • Increment the "distance" variable after processing each level of BFS.

7. Finally, return the maximum "distance" value, which represents the maximum distance from water to land.

During an interview, it is crucial to understand the problem's requirements and constraints. Start by clarifying any ambiguities and considering edge cases. In this problem, it is essential to discuss what the grid represents (water and land) and how the distance is measured (Manhattan distance).

Once you have a clear understanding of the problem, it's essential to come up with an efficient algorithm. BFS is a common choice for problems involving grid-based traversal. Explain the BFS approach, its time and space complexity, and how it can solve the problem optimally. In this problem, the BFS approach helps find the maximum distance from water to land in linear time complexity.

After explaining the algorithm, proceed to implement the code, adding detailed comments for better understanding and maintainability. Remember to discuss the space and time complexity of the code.

Finally, run some test cases to verify the correctness of the solution and perform further optimizations if needed.